Chapter 7, Problem 8P

(a)

To determine

Find the magnetic field intensity $H$ at the point $\left(0,0,5\right)$ due to side 2 of the triangular loop.

Answer to Problem 8P

The magnetic field intensity $H$ at the point $\left(0,0,5\right)$ due to side 2 of the triangular loop is $\left(27.37a_x+27.37a_y+10.95a_z\right)\text{mA}/\text{m}$.

Explanation of Solution

Calculation:

Refer to given Figure in the textbook.

Write the expression to calculate the magnetic field intensity for a straight current carrying conductor.

$H_{\text{side-2}}=\frac{I}{4\pi \rho }\left(\cos {\alpha }_2-\cos {\alpha }_1\right)a_{\varphi }$        (1)

Here,

$I$ is the value of current.

From the given Figure, at point $\left(0,0,5\right)$ due to side 2 is,

$\begin{array}{c}\rho =\sqrt{{\left(0-1\right)}^2+{\left(0-1\right)}^2+{\left(5-0\right)}^2}\\ =\sqrt{1+1+25}\\ =\sqrt{27}\end{array}$

To find $\cos {\alpha }_1$:

$\begin{array}{c}\cos {\alpha }_1=-\frac{2}{\sqrt{2}\left(\sqrt{29}\right)}\\ =-\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}\left(\sqrt{29}\right)}\\ =-\frac{\sqrt{2}}{\sqrt{29}}\\ =-\sqrt{\frac{2}{29}}\end{array}$

To find $\cos {\alpha }_2$:

$\cos {\alpha }_2=0$

Write the general expression to calculate the direction of the magnetic field intensity.

$a_{\varphi }=a_l\times a_{\rho }$        (2)

Here,

$a_l$ is the unit vector along the line current and

$a_{\rho }$ is the unit vector along the perpendicular line to the field point.

For the given conductor the direction of the field is,

$\begin{array}{l}{a}_l=\frac{-a_x+a_y}{\sqrt{2}}\\ a_{\rho }=\frac{-a_x-a_y+5a_z}{\sqrt{27}}\end{array}$

Substitute $\frac{-a_x+a_y}{\sqrt{2}}$ for $a_l$ and $\frac{-a_x-a_y+5a_z}{\sqrt{27}}$ for $a_{\rho }$ in Equation (2).

$\begin{array}{c}{a}_{\varphi }=\left(\frac{-a_x+a_y}{\sqrt{2}}\right)\times \left(\frac{-a_x-a_y+5a_z}{\sqrt{27}}\right)\\ =\frac{5a_x+5a_y+2a_z}{\sqrt{54}}\end{array}$

Substitute $\frac{5a_x+5a_y+2a_z}{\sqrt{54}}$ for $a_{\varphi }$, $10\text{A}$ for $I$, $\sqrt{27}$ for $\rho$, $-\sqrt{\frac{2}{29}}$ for $\cos {\alpha }_1$ and $0$ for $\cos {\alpha }_2$ in Equation (1).

$\begin{array}{c}{H}_{\text{side-2}}=\frac{\left(10\right)}{4\pi \left(\sqrt{27}\right)}\left(0-\left(-\sqrt{\frac{2}{29}}\right)\right)\left(\frac{5a_x+5a_y+2a_z}{\sqrt{54}}\right)\\ =\frac{10}{4\pi \left(\sqrt{27}\right)}\left(\frac{\sqrt{2}}{\sqrt{29}}\right)\left(\frac{1}{\sqrt{54}}\right)\left(5a_x+5a_y+2a_z\right)\\ =\left[\left(27.37\times {10}^{-3}\right)a_x+\left(27.37\times {10}^{-3}\right)a_y+\left(10.95\times {10}^{-3}\right)a_z\right]\text{A}/\text{m}\\ =\left(27.37a_x+27.37a_y+10.95a_z\right)\text{mA}/\text{m}\end{array}$

Conclusion:

Thus, the magnetic field intensity $H$ at the point $\left(0,0,5\right)$ due to side 2 of the triangular loop is $\left(27.37a_x+27.37a_y+10.95a_z\right)\text{mA}/\text{m}$.

(b)

To determine

Find the magnetic field intensity $H$ at the point $\left(0,0,5\right)$ due to the entire loop of the triangular loop.

Answer to Problem 8P

The magnetic field intensity $H$ at the point $\left(0,0,5\right)$ due to the entire loop of the triangular loop is $\left(-3.26a_x-1.1a_y+10.95a_z\right)\text{mA}/\text{m}$.

Explanation of Solution

Calculation:

Write the expression to calculate the magnetic field intensity for the entire loop of a triangle.

$H=H_{\text{side-1}}+H_{\text{side-2}}+H_{\text{side-3}}$        (3)

Find the magnetic field intensity due to the side 1:

Write the expression to calculate the magnetic field intensity for a straight current carrying conductor.

$H_{\text{side-1}}=\frac{I}{4\pi \rho }\left(\cos {\alpha }_2-\cos {\alpha }_1\right)a_{\varphi }$        (4)

From the given Figure, at point $\left(0,0,5\right)$ due to side 1 is,

$\rho =5$

$\cos {\alpha }_1=\cos 90°=0$

$\begin{array}{c}\cos {\alpha }_2=\frac{2}{\sqrt{{\left(5\right)}^2+{\left(2\right)}^2}}\\ =\frac{2}{\sqrt{25+4}}\\ =\frac{2}{\sqrt{29}}\end{array}$

For the given conductor the direction of the field is,

$\begin{array}{l}{a}_l=a_x\\ a_{\rho }=a_z\end{array}$

Substitute $a_x$ for $a_l$ and $a_z$ for $a_{\rho }$ in Equation (2).

$\begin{array}{c}{a}_{\varphi }=a_x\times a_z\\ =-a_y\end{array}$

Substitute $-a_y$ for $a_{\varphi }$, $10\text{A}$ for $I$, $5$ for $\rho$, $0$ for $\cos {\alpha }_1$ and $\frac{2}{\sqrt{29}}$ for $\cos {\alpha }_2$ in Equation (4).

$\begin{array}{c}{H}_{\text{side-1}}=\frac{\left(10\right)}{4\pi \left(5\right)}\left(\frac{2}{\sqrt{29}}-0\right)\left(-a_y\right)\\ =\frac{1}{2\pi }\left(\frac{2}{\sqrt{29}}\right)\left(-a_y\right)\\ =-\left(59.1\times {10}^{-3}\right)a_y\text{A}/\text{m}\\ =-59.1a_y\text{mA}/\text{m}\end{array}$

Find the magnetic field intensity due to the side 3:

Write the expression to calculate the magnetic field intensity for a straight current carrying conductor.

$H_{\text{side-3}}=\frac{I}{4\pi \rho }\left(\cos {\alpha }_2-\cos {\alpha }_1\right)a_{\varphi }$        (5)

From the given Figure and point $\left(0,0,5\right)$,

$\rho =5$

$\cos {\alpha }_1=\cos 90°=0$

$\begin{array}{c}\cos {\alpha }_2=\frac{\sqrt{2}}{\sqrt{{\left(5\right)}^2+{\left(\sqrt{2}\right)}^2}}\\ =\frac{\sqrt{2}}{\sqrt{25+2}}\\ =\frac{\sqrt{2}}{\sqrt{27}}\\ =\sqrt{\frac{2}{27}}\end{array}$

For the given conductor the direction of the field is,

$\begin{array}{l}{a}_l=\frac{-a_x-a_y}{\sqrt{2}}\\ a_{\rho }=a_z\end{array}$

Substitute $\frac{-a_x-a_y}{\sqrt{2}}$ for $a_l$ and $a_z$ for $a_{\rho }$ in Equation (2).

$\begin{array}{c}{a}_{\varphi }=\left(\frac{-a_x-a_y}{\sqrt{2}}\right)\times a_z\\ =\frac{-a_x+a_y}{\sqrt{2}}\end{array}$

Substitute $\frac{-a_x+a_y}{\sqrt{2}}$ for $a_{\varphi }$, $10\text{A}$ for $I$, $5$ for $\rho$, $0$ for $\cos {\alpha }_1$ and $\sqrt{\frac{2}{27}}$ for $\cos {\alpha }_2$ in Equation (5).

$\begin{array}{c}{H}_{\text{side-3}}=\frac{\left(10\right)}{4\pi \left(5\right)}\left(\sqrt{\frac{2}{27}}-0\right)\left(\frac{-a_x+a_y}{\sqrt{2}}\right)\\ =\frac{10}{20\pi }\left(\sqrt{\frac{2}{27}}\right)\left(\frac{1}{\sqrt{2}}\right)\left(-a_x+a_y\right)\\ =-\left(30.63\times {10}^{-3}\right)a_x+\left(30.63\times {10}^{-3}\right)a_y\text{A}/\text{m}\\ =\left(-30.63a_x+30.63a_y\right)\text{mA}/\text{m}\end{array}$

Substitute $-59.1a_y\text{mA}/\text{m}$ for $H_{\text{side-1}}$, $\left(27.37a_x+27.37a_y+10.95a_z\right)\text{mA}/\text{m}$ for $H_{\text{side-2}}$ and $\left(-30.63a_x+30.63a_y\right)\text{mA}/\text{m}$ for $H_{\text{side-3}}$ in Equation (3).

$\begin{array}{c}H=\left[\begin{array}{c}-59.1a_y\text{mA}/\text{m}+\left(27.37a_x+27.37a_y+10.95a_z\right)\text{mA}/\text{m}\\ +\left(-30.63a_x+30.63a_y\right)\text{mA}/\text{m}\end{array}\right]\\ =\left(-59.1a_y+27.37a_x+27.37a_y+10.95a_z-30.63a_x+30.63a_y\right)\text{mA}/\text{m}\\ =\left(-3.26a_x-1.1a_y+10.95a_z\right)\text{mA}/\text{m}\end{array}$

Conclusion:

Thus, the magnetic field intensity $H$ at the point $\left(0,0,5\right)$ due to the entire loop of the triangular loop is $\left(-3.26a_x-1.1a_y+10.95a_z\right)\text{mA}/\text{m}$.